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HSC 2015 MX1 Marathon (archive) (4 Viewers)

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InteGrand

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Re: HSC 2015 3U Marathon

davidgoes4wce said:
Curious to also know has a 'First Principles' question ever popped up in previous years' exam papers for MX1? 3 unit?

One of the students who I chat to says that their school has not covered it and the teachers have not put it into their school curriculum
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It has. See page 10 Q.7 (a) (i) of the 2009 HSC: http://www.boardofstudies.nsw.edu.a...doc/2009-hsc-exam-mathematics-extension-1.pdf.

And I'm pretty sure the teachers should have at least covered first principles differentiation in 2U.
 
Speed6

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Re: HSC 2015 3U Marathon

Solve the equation:

 
leehuan

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Re: HSC 2015 3U Marathon

Assuming degrees.

90deg, 270deg

Shouldn't this be in the 2U thread?


Edit: Qn done above in rads
 
Speed6

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Re: HSC 2015 3U Marathon

leehuan said:
Assuming degrees.

90deg, 270deg

Shouldn't this be in the 2U thread?


Edit: Qn done above in rads
Click to expand...
Hmm I think this is leaning towards 3U???

Simple question but it can also be considered 2U..
 
leehuan

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Re: HSC 2015 3U Marathon

2U are expected to know how to change a domain and take positive and negative roots so I wouldn't consider this harder 2U for the 3U student.

Something at the 3U exclusive level would be something like sin(x) + sqrt(3)cos(x) = 1, which uses the auxiliary angle method.

Harder 2U would be something rigorous involving tan^2(x) and sec^2(x) and cos^2(x) and etc all in the one go, which requires extensive use of Pythagorean identities and the ratio identities of tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x)
 
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Re: HSC 2015 3U Marathon

Speed6 said:
Hmm I think this is leaning towards 3U???

Simple question but it can also be considered 2U..
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It depends on how you approach the question

Ekman did it using a 3 unit method (i.e. Double angle)

He used the identity: cos(2x) = 1 - 2sin^2(x)
 
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Re: HSC 2015 3U Marathon

Fitzpatricks book definies the locus "A problem to be solved by eliminating the parameter from some parametric equations (ususally x(t)), y(t)) to form a new Cartesian equation (y=f(x))
 
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SammyT123

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Re: HSC 2015 3U Marathon

Can everyone see the latex on tapatalk? I seem to be having issues rendering it.


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HSC 2015 3U Marathon

davidgoes4wce said:
I dont see Latex on yout tapatalk or the message you posted
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I mean for previous questions , I can't see any latex at all when I'm on my phone


Sent from my iPhone using Tapatalk
 
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SammyT123

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Re: HSC 2015 3U Marathon

davidgoes4wce said:
Fitzpatricks book definies the locus "A problem to be solved by eliminating the parameter from some parametric equations (ususally x(t)), y(t)) to form a new Cartesian equation (y=f(x))
Click to expand...
Eliminate p by making it the subject .
X=...
Y=...
And link the equations


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